I am trying to do the stated problem in Hatcher:
Show $H_1(X,A) = 0$ iff $H_1(A) \to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.
Now I have reduced the problem to showing that $i_\ast : H_0(A) \to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$\ldots \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_\ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_\ast$ is not injective. Then there is a $\tau \in C_0(A)$ such that $[\tau \circ i] = 0$ but $[\tau] \neq 0$. That is to say, $\tau \circ i = \partial(\sigma)$ for some $\sigma \in C_1(X)$ but $\tau$ is not the boundary of any $\sigma'\in C_1(A)$. However I'm confused because to me the only way for $\tau \circ i$ to be the boundary of a singular $1$ - simplex $\sigma$ in $X$ is if $\sigma$ is a loop. What's wrong here?